Original JEE Main and CUET-style questions on trigonometry and inverse trigonometric functions, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 11, Trigonometric Functions · Class 12, Inverse Trigonometric Functions.
The number of solutions of \(\sin 2x = \cos x\) in \([0, 2\pi]\) is
\(2\sin x\cos x - \cos x = 0\), so \(\cos x\,(2\sin x - 1) = 0\).
\(\cos x = 0\): \(x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}\). \(\sin x = \tfrac12\): \(x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}\). Four solutions.
Trap: Dividing both sides by \(\cos x\) throws away the two solutions where \(\cos x = 0\).
The value of \(\sin^{-1}\!\left(\sin \dfrac{5\pi}{6}\right)\) is
\(\sin^{-1}\) returns a value in \(\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\). Since \(\sin \tfrac{5\pi}{6} = \tfrac12\), the answer is \(\sin^{-1}\tfrac12 = \tfrac{\pi}{6}\).
Trap: \(\sin^{-1}(\sin\theta) = \theta\) only when \(\theta\) is already in \([-\pi/2, \pi/2]\); \(5\pi/6\) is not.
Find the greatest value of \(7\sin x - 24\cos x + 5\).
\(a\sin x + b\cos x\) lies between \(-\sqrt{a^2 + b^2}\) and \(\sqrt{a^2 + b^2}\). Here \(\sqrt{49 + 576} = 25\), so the greatest value is \(25 + 5 = 30\).
Trap: Adding the largest values of the two terms separately (7 + 24 + 5 = 36) is wrong: \(\sin x = 1\) and \(\cos x = -1\) never happen together.
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